---
title: "Propped cantilever with UDL by integration"
subject: "Strength of Materials"
chapter: 7
key: "semester_4/strength_of_materials_mechanical:propped-cantilever-udl-prop-reaction"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/propped-cantilever-udl-prop-reaction
site: BCT Engineering
---

# Propped cantilever with UDL by integration

Strength of Materials · Chapter 7: Deflection of Beams by Integration Method

- Asked in: year not given, 8 marks
- Video: [Propped Cantilever beam with Uniformly Distributed Load ( UDL ) for the Full Span](https://www.youtube.com/watch?v=8cUpr-Sdo7U) — Stan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/propped-cantilever-udl-prop-reaction
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-7

## Question

A cantilever of span 4 m fixed at A carries a uniformly distributed load of 20 kN/m over its whole length and is propped at the free end B so that B does not deflect. EI = 16 000 kN m^2 is constant. Using the double integration method find (a) the prop reaction, (b) the fixed-end reactions and moment, (c) the position and value of the maximum deflection, (d) the point of contraflexure.

## Short answer

Prop reaction = 30 kN. Fixed-end vertical reaction = 50 kN, hogging moment = 40 kN·m. Maximum deflection = 1.73 mm downward at 2.31 m from fixed end. Point of contraflexure at 1 m from fixed end.

## Answer

#### Statically indeterminate beam
```text
 ||A===================B
 ||  w = 20 kN/m       ^ R_B (prop)
 ||<------ 4 m ------>
```

Let $R_B$ be the prop reaction. Take the origin at A, $x$ towards B, sagging moment positive and $y$ upward.

Let $R_A = wL - R_B$ be the vertical reaction at the fixed end A, and $M_A$ the hogging fixed-end moment. The bending moment at a section $x$ from A gives

$$
EI\,y'' = R_Ax - M_A - \frac{wx^2}{2}
$$

$$
EI\,y' = \frac{R_Ax^2}{2} - M_Ax - \frac{wx^3}{6} \quad (C_1=0\text{ since } y'=0\text{ at }x=0)
$$

$$
EI\,y = \frac{R_Ax^3}{6} - \frac{M_Ax^2}{2} - \frac{wx^4}{24} \quad (C_2=0\text{ since } y=0\text{ at }x=0)
$$

#### (a) Prop reaction
Conditions: $y = 0$ at $x = L$ (prop), and $M(L) = 0$ at the free end B (no moment is applied there):

$$
R_AL - M_A - \frac{wL^2}{2} = 0 \;\Rightarrow\; M_A = R_AL - \frac{wL^2}{2}
$$

$$
EI\,y_B = \frac{R_AL^3}{6} - \frac{M_AL^2}{2} - \frac{wL^4}{24} = 0
$$

Substituting $M_A$: $\dfrac{R_AL^3}{6} - \dfrac{R_AL^3}{2} + \dfrac{wL^4}{4} - \dfrac{wL^4}{24} = 0$, hence $\dfrac{R_AL^3}{3} = \dfrac{5wL^4}{24}$, so $R_A = \dfrac{5wL}{8}$.

$$
R_A = \frac{5\times20\times4}{8} = 50\ \text{kN}, \qquad R_B = wL - R_A = 80 - 50 = 30\ \text{kN}
$$

#### (b) Fixed-end reactions and moment
$R_A = 50$ kN (upward), and

$$
M_A = R_AL - \frac{wL^2}{2} = 50\times4 - 160 = 40\ \text{kN m (hogging)} \;\left(=\frac{wL^2}{8}\right)
$$

Check, moments about A: $30\times4 - 80\times2 + 40 = 0$. ✓

#### (c) Maximum deflection ($y'=0$)
$$
EI\,y' = 25x^2 - 40x - \frac{20x^3}{6} = 0 \;\Rightarrow\; 3.333x^2 - 25x + 40 = 0
$$

$$
x = \frac{25 - \sqrt{625 - 533.3}}{6.667} = \frac{25 - 9.57}{6.667} = 2.314\ \text{m}\ (=0.5785L\text{ from A, i.e. }0.4215L\text{ from the prop})
$$

$$
EI\,y = \frac{50(2.314)^3}{6} - \frac{40(2.314)^2}{2} - \frac{20(2.314)^4}{24} = 103.2 - 107.1 - 23.9 = -27.7
$$

$$
y_{max} = \frac{-27.7}{16000} = -1.73\times10^{-3}\ \text{m} = 1.73\ \text{mm downward}
$$

#### (d) Point of contraflexure
$M(x) = 50x - 40 - 10x^2 = 0 \Rightarrow x^2 - 5x + 4 = 0 \Rightarrow x = 1\ \text{m}$ or $4\ \text{m}$. The root $x = 4$ m is the prop itself, so the point of contraflexure is at $x = 1$ m from A. The maximum sagging moment is where $V = 0$, at $x = 50/20 = 2.5$ m: $M = 50(2.5) - 40 - 10(6.25) = 22.5$ kN m.

**Answer: $R_B = 30$ kN, $R_A = 50$ kN, $M_A = 40$ kN m; $y_{max} = 1.73$ mm at 2.31 m from A; contraflexure at 1 m from A.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why can the beam not be solved by equilibrium alone, and what extra condition is used?
   A. It is statically indeterminate; deflection at the prop is zero
   B. It is statically determinate; slope at the free end is zero
   C. It has a hinge; moment at the fixed end is zero
   D. It is statically indeterminate; the deflection at the fixed end is zero only
   Answer: A. Three unknown reactions exceed the equations of equilibrium, so yB = 0 is added. (Key point: Extra condition)

2. What is the fixed-end vertical reaction RA for a propped cantilever with UDL?
   A. 3wL/8 = 30 kN
   B. 5wL/8 = 50 kN
   C. wL/2 = 40 kN
   D. wL = 80 kN
   Answer: B. Solving yB = 0 with M(L) = 0 gives RA = 5wL/8, and RB = 3wL/8 = 30 kN. (Key point: Prop reaction)

3. What is the fixed-end moment?
   A. 80 kN m hogging, wL²/4
   B. 40 kN m sagging
   C. 40 kN m hogging, wL²/8
   D. 160 kN m hogging, wL²/2
   Answer: C. MA = RA L - wL²/2 = 200 - 160 = 40 kN m. (Key point: Fixed-end moment)

4. Where is the point of contraflexure?
   A. 2.5 m from A, where shear is zero
   B. 4 m from A, at the prop
   C. 2 m from A, at mid-span
   D. 1 m from A, from 50x - 40 - 10x² = 0
   Answer: D. Roots are x = 1 and 4; x = 4 is the prop end, so the sign change inside the span is at 1 m. (Key point: Contraflexure)

## Related in Strength of Materials

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